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Date May Example question Marks available 3 Reference code EXM.2.AHL.TZ0.20
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Deduce Question number 20 Adapted from N/A

Question

Let γ=1+i32.

The matrix A is defined by A (γ101γ).

Deduce that

Show that γ2=γ1.

[2]
a.ii.

Hence find the value of (1γ)6.

[4]
a.iii.

A3 = –I.

[3]
c.i.

A–1 = IA.

[2]
c.ii.

Markscheme

METHOD 1

as γ is a root of z2z+1=0 then γ2γ+1=0       M1R1

γ2=γ1     AG

Note: Award M1 for the use of z2z+1=0 in any way.

Award R1 for a correct reasoned approach.

 

METHOD 2

γ2=1+i32       M1

γ1=1+i321=1+i32        A1 

 

[2 marks]

a.ii.

METHOD 1

(1γ)6=(γ2)6       (M1)

               =(γ)12     A1

               =(γ3)4       (M1)

               =(1)4

               =1     A1

 

METHOD 2

(1γ)6

=16γ+15γ220γ3+15γ46γ5+γ6       M1A1

Note: Award M1 for attempt at binomial expansion.

use of any previous result e.g. =16γ+15γ2+2015γ+6γ2+1       M1

=1     A1

Note: As the question uses the word ‘hence’, other methods that do not use previous results are awarded no marks.

 

[4 marks]

a.iii.

A2 = A I  

A3 = A2 – A      M1A1           

          = AIA         A1           

          = I           AG

Note: Allow other valid methods.

[3 marks]

c.i.

I = A A2

A–1 = A–1AA–1A2        M1A1

⇒ A–1 = I A         AG

Note: Allow other valid methods.

[2 marks]

c.ii.

Examiners report

[N/A]
a.ii.
[N/A]
a.iii.
[N/A]
c.i.
[N/A]
c.ii.

Syllabus sections

Topic 1—Number and algebra » AHL 1.12—Complex numbers introduction
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